let the region r be the area enclosed by the function f(x)=x², the horizontal line y = 6, and the y - axis…

Explanation:

Step1: Find intersection points

Set $y = x^{2}$ and $y = 6$. Then $x^{2}=6$, so $x=\sqrt{6}$ (since we are in the first - quadrant considering the region bounded by the $y$-axis). The limits of integration for $x$ are from $x = 0$ to $x=\sqrt{6}$.

Step2: Determine the leg length of the isosceles right - triangle

The length of a leg $l$ of the isosceles right - triangle cross - section perpendicular to the $x$-axis is given by the vertical distance in the region $R$. So $l=6 - x^{2}$.

Step3: Find the area formula for the isosceles right - triangle

The area formula for an isosceles right - triangle with leg length $l$ is $A=\frac{1}{2}l^{2}$. Substituting $l = 6 - x^{2}$, we get $A(x)=\frac{1}{2}(6 - x^{2})^{2}=\frac{1}{2}(36-12x^{2}+x^{4})$.

Step4: Calculate the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = a$ to $x = b$ is $V=\int_{a}^{b}A(x)dx$. Here, $a = 0$, $b=\sqrt{6}$, and $A(x)=\frac{1}{2}(36-12x^{2}+x^{4})$. So $V=\int_{0}^{\sqrt{6}}\frac{1}{2}(36 - 12x^{2}+x^{4})dx$. [ \begin{align*} V&=\frac{1}{2}\int_{0}^{\sqrt{6}}(36-12x^{2}+x^{4})dx\ &=\frac{1}{2}\left[36x-12\times\frac{x^{3}}{3}+\frac{x^{5}}{5}\right]_{0}^{\sqrt{6}}\ &=\frac{1}{2}\left(36\sqrt{6}-4\times(\sqrt{6})^{3}+\frac{(\sqrt{6})^{5}}{5}\right)\ &=\frac{1}{2}\left(36\sqrt{6}-4\times6\sqrt{6}+\frac{6^{2}\times\sqrt{6}}{5}\right)\ &=\frac{1}{2}\sqrt{6}\left(36 - 24+\frac{36}{5}\right)\ &=\frac{1}{2}\sqrt{6}\left(12+\frac{36}{5}\right)\ &=\frac{1}{2}\sqrt{6}\left(\frac{60 + 36}{5}\right)\ &=\frac{1}{2}\sqrt{6}\times\frac{96}{5}\ &=\frac{48\sqrt{6}}{5}\approx23.510 \end{align*} ]

Answer:

$23.510$